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Alphametic Problem Solving Strategies

Alphametics are a type of verbal arithmetic brainteaser. They look easy to solve, but they can prove to be a real head scratcher if you just take a trial-and-error approach.

Two basic rules guide solving alphametics.

Each letter must be represented by a different digit. If the letter is used more than once, it must be represented by the same digit.

Once you substitute digits for all your letters, you must wind up with an accurate addition problem.

Before you start solving these brainteasers, take a minute to remember the Additive Identity and the Commutative Properties of addition. Think about how these properties may be represented in an alphametic puzzle.

For example, the Additive Property – the rule that any number plus zero always equals itself – would look like:

A + B = A

That sure seems easy, until your numbers get bigger and you start to carry groups of ten into your next column.

Let’s see a problem solving strategy in action.

Now that we know we’ll be carrying a group of tens, we know three basic facts about this problem:

A + C >= 10

B + 1 = D

B cannot equal 0 or 9

At this point, we can plug some numbers in to see what might satisfy this equation. We’ll start by brainstorming any set of facts that equal 10, 11, 12, … 17. (We know the sum of the one’s column cannot equal 18, because the only two single digits that equal that sum are 9 and 9. A and C must be different digits.)

If we randomly choose A = 7 and C = 5 we see:

We’ve successfully solved this problem! We could have, however, chosen different values for A and C and still gotten a correct solution. This particulare problem has 18 different answers that work.

Let’s take a look at a harder alphametic problem:

Before we start guessing numbers, let’s analyze the one’s and ten’s column.

B + D = E

B + C = E

Since each letter must be represented by a different digit, the only way both of these statements can be true is if we carry a group of tens into the B + C column. Knowing that means:

C + 1 = D

Next, we look at the hundreds column. The only way A = C is if we carry over to this column, also. We now have a second rule we must remember:

A + 1 = C

At this point, we have enough information to make an educated guess on how to solve this puzzle. We can begin by randomly choosing A=1, C=2, and D=3.

Since B + 3 must be equal to or greater than 10 and E cannot equal 1, 2, or 3 (because those are the digits we’re using for A, C, and D), the only possible answer is that B = 7.

Had we started by making A equal a number other than one, we would still be able to solve the puzzle using the same basic rules.